Circuits Practice Problems for AP Physics B & C (page 2)

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By — McGraw-Hill Professional
Updated on Feb 12, 2011

Solutions to Practice Problems

  1. A—On one hand, you could use a V-I-R chart to calculate the voltage or current for each resistor, then use P = IV, I2R, or V2/R to find power. On the other hand, there's a quick way to reason through this one. Voltage changes across the 100 Ω resistor, then again across the parallel combination. Because the 100 Ω resistor has a bigger resistance than the parallel combination, the voltage across it is larger as well. Now consider each resistor individually. By power = V2/R, the 100 Ω resistor has both the biggest voltage and the smallest resistance, giving it the most power.
  2. A—The energy stored by a capacitor is ½CV2. By powering a car, this electrical energy is converted into mechanical work, equal to force times parallel displacement. Solve for displacement, you get 36 m.
  3. C—To use Ohm's law here, simplify the circuit to a 10 V battery with the 10 Ω equivalent resistance. We can use Ohm's law for the entire circuit to find that 1.0 A is the total current. Because all the resistors are in series, this 1.0 A flows through each resistor, including the 2 Ω resistor.
  4. D—First, simplify the circuit to find the equivalent capacitance. The parallel capacitors add to 6 μF. Then the two series capacitors combine to 3 μF. So we end up with 9 V across a 3-μF equivalent capacitance. By the basic equation for capacitors, Q = CV, the charge stored on these capacitors is 27 μC.
  5. A—An ammeter is placed in series with other circuit components. In order for the ammeter not to itself resist current and change the total current in the circuit, you want the ammeter to have as little resistance as possible—in the ideal case, zero resistance. But a voltmeter is placed in parallel with other circuit components. If the voltmeter has a low resistance, then current will flow through the voltmeter instead of through the rest of the circuit. Therefore, you want it to have as high a resistance as possible, so the voltmeter won't affect the circuit being measured.
  7. (a) Combine each of the sets of parallel resistors first. You get 120 Ω for the first set, 222 Ω for the second set, as shown in the diagram below. These two equivalent resistances add as series resistors to get a total resistance of 342 Ω.

    Practice Problem

    (b) Now that we've found the total resistance and we were given the total voltage, just use Ohm's law to find the total current to be 0.026 A (also known as 26 mA).

    (c) and (d) should be solved together using a V-I-R chart. Start by going back one step to when we began to simplify the circuit: 9-V battery, a 120 Ω combination, and a 222 Ω combination, shown above. The 26-mA current flows through each of these… so use V = IR to get the voltage of each: 3.1 V and 5.8 V, respectively.


    Now go back to the original circuit. We know that voltage is the same across parallel resistors. So both the 200-#937; and 300-Ω resistors have a 3.1-V voltage across them. Use Ohm's law to find that 16 mA goes through the 200-Ω resistor, and 10 mA through the 300 Ω. Similarly, both the 400-Ω and 500-Ω resistors must have 5.8 V across them. We get 15 mA and 12 mA, respectively. Checking these answers for reasonability: the total voltage adds to 8.9 V, or close enough to 9.0 V with rounding. The current through each set of parallel resistors adds to just about 26 mA, as we expect.

    (e) Start by looking at the circuit as a whole. When we remove the 500 Ω resistor, we actually increase the overall resistance of the circuit because we have made it more difficult for current to flow by removing a parallel path. The total voltage of the circuit is provided by the battery, which provides 9.0 V no matter what it's hooked up to. So by Ohm's law, if total voltage stays the same while total resistance increases, total current must decrease from 26 mA.

    Okay, now look at the first set of parallel resistors. Their equivalent resistance doesn't change, yet the total current running through them decreases, as discussed above. Therefore, the voltage across each resistor decreases, and the current through each decreases as well.

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